$2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$ are $APs$? If they form an $AP$,find the common difference $d$ and write three more terms.

(A) Given sequence: $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$
To check if it is an $AP$,we calculate the difference between consecutive terms:
$a_{2} - a_{1} = \frac{5}{2} - 2 = \frac{1}{2}$
$a_{3} - a_{2} = 3 - \frac{5}{2} = \frac{1}{2}$
$a_{4} - a_{3} = \frac{7}{2} - 3 = \frac{1}{2}$
Since the difference $a_{k+1} - a_{k}$ is constant,the sequence is an $AP$ with common difference $d = \frac{1}{2}$.
The next three terms are:
$a_{5} = \frac{7}{2} + \frac{1}{2} = 4$
$a_{6} = 4 + \frac{1}{2} = \frac{9}{2}$
$a_{7} = \frac{9}{2} + \frac{1}{2} = 5$